Integrand size = 35, antiderivative size = 506 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {(a-b) \sqrt {a+b} \left (15 A b^2+8 a^2 (2 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^3 b d}-\frac {\sqrt {a+b} \left (10 a A b-15 A b^2-8 a^2 (2 A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^3 d}+\frac {b \sqrt {a+b} \left (5 A b^2+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{8 a^4 d}+\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^3 d}-\frac {5 A b \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a^2 d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 a d} \]
1/24*(a-b)*(15*A*b^2+8*a^2*(2*A+3*C))*cot(d*x+c)*EllipticE((a+b*sec(d*x+c) )^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+ b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/b/d-1/24*(10*a*A*b-15*A*b^2- 8*a^2*(2*A+3*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),( (a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec( d*x+c))/(a-b))^(1/2)/a^3/d+1/8*b*(5*A*b^2+4*a^2*(A+2*C))*cot(d*x+c)*Ellipt icPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b) ^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/ d+1/24*(15*A*b^2+8*a^2*(2*A+3*C))*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^3/d- 5/12*A*b*cos(d*x+c)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^2/d+1/3*A*cos(d*x+ c)^2*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d
Leaf count is larger than twice the leaf count of optimal. \(1352\) vs. \(2(506)=1012\).
Time = 21.08 (sec) , antiderivative size = 1352, normalized size of antiderivative = 2.67 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx =\text {Too large to display} \]
((b + a*Cos[c + d*x])*Sec[c + d*x]*((A*Sin[c + d*x])/(12*a) - (5*A*b*Sin[2 *(c + d*x)])/(24*a^2) + (A*Sin[3*(c + d*x)])/(12*a)))/(d*Sqrt[a + b*Sec[c + d*x]]) + (Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^ 2)/(1 + Tan[(c + d*x)/2]^2)]*(-16*a^3*A*Tan[(c + d*x)/2] - 16*a^2*A*b*Tan[ (c + d*x)/2] - 15*a*A*b^2*Tan[(c + d*x)/2] - 15*A*b^3*Tan[(c + d*x)/2] - 2 4*a^3*C*Tan[(c + d*x)/2] - 24*a^2*b*C*Tan[(c + d*x)/2] + 32*a^3*A*Tan[(c + d*x)/2]^3 + 30*a*A*b^2*Tan[(c + d*x)/2]^3 + 48*a^3*C*Tan[(c + d*x)/2]^3 - 16*a^3*A*Tan[(c + d*x)/2]^5 + 16*a^2*A*b*Tan[(c + d*x)/2]^5 - 15*a*A*b^2* Tan[(c + d*x)/2]^5 + 15*A*b^3*Tan[(c + d*x)/2]^5 - 24*a^3*C*Tan[(c + d*x)/ 2]^5 + 24*a^2*b*C*Tan[(c + d*x)/2]^5 + 24*a^2*A*b*EllipticPi[-1, ArcSin[Ta n[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*A*b^3*Ellipt icPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x) /2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 48*a^2*b*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sq rt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 24*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*A*b^3*...
Time = 2.32 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4593, 27, 3042, 4592, 27, 3042, 4592, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4593 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-3 A b \sec ^2(c+d x)-2 a (2 A+3 C) \sec (c+d x)+5 A b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-3 A b \sec ^2(c+d x)-2 a (2 A+3 C) \sec (c+d x)+5 A b\right )}{\sqrt {a+b \sec (c+d x)}}dx}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (8 (2 A+3 C) a^2+2 A b \sec (c+d x) a+15 A b^2-5 A b^2 \sec ^2(c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{2 a}}{6 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (8 (2 A+3 C) a^2+2 A b \sec (c+d x) a+15 A b^2-5 A b^2 \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\int \frac {8 (2 A+3 C) a^2+2 A b \csc \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^2-5 A b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {10 a A \sec (c+d x) b^2+\left (8 (2 A+3 C) a^2+15 A b^2\right ) \sec ^2(c+d x) b+3 \left (4 (A+2 C) a^2+5 A b^2\right ) b}{2 \sqrt {a+b \sec (c+d x)}}dx}{a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {10 a A \sec (c+d x) b^2+\left (8 (2 A+3 C) a^2+15 A b^2\right ) \sec ^2(c+d x) b+3 \left (4 (A+2 C) a^2+5 A b^2\right ) b}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {10 a A \csc \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (8 (2 A+3 C) a^2+15 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b+3 \left (4 (A+2 C) a^2+5 A b^2\right ) b}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {3 b \left (4 (A+2 C) a^2+5 A b^2\right )+\left (10 a A b^2-b \left (8 (2 A+3 C) a^2+15 A b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {3 b \left (4 (A+2 C) a^2+5 A b^2\right )+\left (10 a A b^2-b \left (8 (2 A+3 C) a^2+15 A b^2\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+3 b \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {3 b \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 b \sqrt {a+b} \left (4 a^2 (A+2 C)+5 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {6 b \sqrt {a+b} \left (4 a^2 (A+2 C)+5 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d}-\frac {\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\frac {2 \sqrt {a+b} \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {6 b \sqrt {a+b} \left (4 a^2 (A+2 C)+5 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{2 a}}{4 a}}{6 a}\) |
(A*Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a*d) - ((5*A*b *Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*a*d) - (-1/2*((-2* (a - b)*Sqrt[a + b]*(15*A*b^2 + 8*a^2*(2*A + 3*C))*Cot[c + d*x]*EllipticE[ ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (2*Sqrt[a + b]*(10*a*A*b - 15*A*b^2 - 8*a^2*(2*A + 3*C))*Cot[c + d*x]*Ell ipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt [(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/ d - (6*b*Sqrt[a + b]*(5*A*b^2 + 4*a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[( a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) ])/(a*d))/a + ((15*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Sec[c + d*x]]*Sin [c + d*x])/(a*d))/(4*a))/(6*a)
3.8.41.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[( -A)*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && NeQ[a^2 - b^2 , 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3151\) vs. \(2(461)=922\).
Time = 7.09 (sec) , antiderivative size = 3152, normalized size of antiderivative = 6.23
1/24/d/a^3*(-30*A*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^( 1/2)*a*b^2*cos(d*x+c)+20*A*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^( 1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x +c)+1))^(1/2)*a*b^2*cos(d*x+c)-48*C*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b) /(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c)) /(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)-15*A*EllipticE(cot(d*x+c)-csc(d*x+ c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co s(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^2*cos(d*x+c)^2+10*A*EllipticF(cot(d*x+ c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a +b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^2*cos(d*x+c)^2-24*C*Ellipti cE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)^2+2 4*C*a^2*b*cos(d*x+c)*sin(d*x+c)-10*A*a*b^2*cos(d*x+c)*sin(d*x+c)-15*A*(cos (d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1 /2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^3-24*C*(cos(d*x +c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)* EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3-16*A*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) *(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b-32*A*EllipticE(c...
\[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]